yes, that's right, you need to differentiate it.
let
$\displaystyle f(x) = 2x^2-12x+14$
when $\displaystyle x = 1$
$\displaystyle f(x) = 2-12+14 = 4$
so the point (1,4) does lie on the curve.
$\displaystyle f'(x) = 4x-12$
$\displaystyle f'(1) = 4-12 = -8$
so the gradient of the curve at (1,4) is -8.
you could also find the equation of the tangent line that touches the curve at (1,4) but i don't think that's what you need to do.